Time AND Work Problems FOR Placement

Meta Description: Master Time and Work problems for campus placements with 20 practice questions, expert tips, and shortcut formulas. Essential guide for TCS, Infosys, Wipro, and all major companies.

Introduction

Time and Work problems are among the most frequently asked questions in campus placement exams of IT giants like TCS, Infosys, Wipro, Accenture, and Cognizant. These questions test your ability to calculate efficiency, completion time, and workforce distribution, skills that mirror real-world project management scenarios.

Understanding Time and Work concepts is crucial because:

High Weightage: 3-5 questions appear in every aptitude test
Quick Solving: Can be solved in 30-45 seconds with the right approach
Conceptual Clarity: Builds foundation for Pipes and Cisterns, Work and Wages topics
Interview Relevance: Often discussed in technical interviews to assess logical thinking

At PapersAdda, we've analyzed thousands of placement papers to bring you the most comprehensive guide with 20 carefully selected practice questions.

20 Practice Questions with Detailed Solutions

Question 1

A can complete a work in 12 days and B can complete the same work in 18 days. In how many days will they complete the work together?

Solution: A's 1 day work = 1/12 B's 1 day work = 1/18 (A + B)'s 1 day work = 1/12 + 1/18 = (3 + 2)/36 = 5/36 Time taken together = 36/5 = 7.2 days

Question 2

A can do a piece of work in 10 days, B in 15 days. They work together for 5 days, then B leaves. How many more days will A take to finish?

Solution: (A + B)'s 1 day work = 1/10 + 1/15 = (3 + 2)/30 = 5/30 = 1/6 Work done in 5 days = 5 × (1/6) = 5/6 Remaining work = 1 - 5/6 = 1/6 A's 1 day work = 1/10 Time for A to complete remaining work = (1/6) ÷ (1/10) = 10/6 = 5/3 = 1.67 days

Question 3

A is twice as good a workman as B. Together they finish the work in 14 days. In how many days can A alone finish the work?

Solution: If A takes x days, B takes 2x days (since A is twice as efficient) A's 1 day work = 1/x, B's 1 day work = 1/2x Together: 1/x + 1/2x = 1/14 3/2x = 1/14 2x = 42, so x = 21

Question 4

12 men can complete a work in 8 days. After 3 days, 4 more men join. In how many days will the remaining work be completed?

Solution: Total work = 12 men × 8 days = 96 man-days Work done in 3 days = 12 × 3 = 36 man-days Remaining work = 96 - 36 = 60 man-days Now total men = 12 + 4 = 16 Days required = 60/16 = 3.75 days

Question 5

A and B together can do a work in 6 days. A alone can do it in 10 days. How long will B alone take?

Solution: (A + B)'s 1 day work = 1/6 A's 1 day work = 1/10 B's 1 day work = 1/6 - 1/10 = (5 - 3)/30 = 2/30 = 1/15 B alone takes 15 days

Question 6

If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, how long will 15 men and 20 boys take?

Solution: Let 1 man's 1 day work = m, 1 boy's 1 day work = b 6m + 8b = 1/10 ... (i) 26m + 48b = 1/2 ... (ii)

Multiply (i) by 6: 36m + 48b = 6/10 = 3/5 Subtract from (ii): -10m = 1/2 - 3/5 = (5-6)/10 = -1/10 m = 1/100

From (i): 6/100 + 8b = 1/10 8b = 1/10 - 6/100 = (10-6)/100 = 4/100 = 1/25 b = 1/200

15 men + 20 boys = 15/100 + 20/200 = 3/20 + 1/10 = (3+2)/20 = 5/20 = 1/4 Time = 4 days

Question 7

A can finish a work in 24 days, B in 9 days, and C in 12 days. B and C start the work but leave after 3 days. How many days will A take to complete the remaining work?

Solution: B's 1 day work = 1/9, C's 1 day work = 1/12 (B + C)'s 1 day work = 1/9 + 1/12 = (4+3)/36 = 7/36 Work done in 3 days = 3 × 7/36 = 7/12 Remaining work = 1 - 7/12 = 5/12 A's 1 day work = 1/24 Time for A = (5/12) ÷ (1/24) = (5/12) × 24 = 10 days

Question 8

20 women can do a work in 16 days, and 16 men can complete the same work in 15 days. What is the ratio between the capacity of a man and a woman?

Solution: 20 women × 16 days = 16 men × 15 days 320 woman-days = 240 man-days 1 man-day = 320/240 woman-days = 4/3 woman-days Ratio of man : woman = 4 : 3

Question 9

A and B can do a piece of work in 72 days; B and C in 120 days; A and C in 90 days. In how many days can A alone do it?

Solution: A + B = 1/72 ... (i) B + C = 1/120 ... (ii) A + C = 1/90 ... (iii)

Adding all three: 2(A + B + C) = 1/72 + 1/120 + 1/90 = (5 + 3 + 4)/360 = 12/360 = 1/30 A + B + C = 1/60

A = (A + B + C) - (B + C) = 1/60 - 1/120 = (2-1)/120 = 1/120 A alone takes 120 days

Question 10

A contractor undertakes to do a piece of work in 40 days. He engages 100 men and after 35 days, he adds 100 more men and completes the work in time. How many days behind schedule would the work have been if he had not engaged the additional men?

Solution: Work done by 100 men in 35 days = 100 × 35 = 3500 man-days Remaining work done by 200 men in 5 days = 200 × 5 = 1000 man-days Total work = 4500 man-days

If only 100 men worked: Days needed = 4500/100 = 45 days Delay = 45 - 40 = 5 days

Question 11

A is 30% more efficient than B. How much time will they take working together to complete a job which A alone could have done in 23 days?

Solution: If B takes x days, A takes x/1.3 days (30% more efficient means 30% less time) x/1.3 = 23, so x = 29.9 ≈ 30 days

A's 1 day work = 1/23, B's 1 day work = 1/30 Together: 1/23 + 1/30 = (30 + 23)/(23×30) = 53/690 Time = 690/53 = 13.02 days ≈ 13 days

Question 12

10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?

Solution: 1 man's 1 day work = 1/100 10 men's 1 day work = 10/100 = 1/10

Let 1 woman's 1 day work = 1/w 10/100 + 15/w = 1/6 15/w = 1/6 - 1/10 = (5-3)/30 = 2/30 = 1/15 w = 225

Question 13

A can do a certain job in 25 days which B alone can do in 20 days. A started the work and was joined by B after 10 days. How many days will the work last from the beginning?

Solution: A's 1 day work = 1/25, B's 1 day work = 1/20 Work done by A in 10 days = 10/25 = 2/5 Remaining work = 3/5

(A + B)'s 1 day work = 1/25 + 1/20 = (4+5)/100 = 9/100 Time to complete remaining work = (3/5) ÷ (9/100) = (3/5) × (100/9) = 60/9 = 20/3 days

Total time = 10 + 20/3 = 50/3 = 16⅔ days

Question 14

4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?

Solution: Let 1 man's 1 day work = m, 1 woman's = w 4m + 6w = 1/8 ... (i) 3m + 7w = 1/10 ... (ii)

Multiply (i) by 3: 12m + 18w = 3/8 Multiply (ii) by 4: 12m + 28w = 4/10 = 2/5

Subtract: 10w = 2/5 - 3/8 = (16-15)/40 = 1/40 w = 1/400

10 women's 1 day work = 10/400 = 1/40 Time = 40 days

Question 15

A, B, and C can complete a work separately in 24, 36, and 48 days respectively. They started together but C left after 4 days and A left 3 days before completion. How many days did the work last?

Solution: Let total days = x A worked for (x-3) days, B worked for x days, C worked for 4 days

(x-3)/24 + x/36 + 4/48 = 1 (x-3)/24 + x/36 + 1/12 = 1 (x-3)/24 + x/36 = 11/12

Multiply by 72: 3(x-3) + 2x = 66 3x - 9 + 2x = 66 5x = 75, x = 15

Question 16

If 5 men or 7 women can earn ₹875 per day, how much would 10 men and 5 women earn per day?

Solution: 5 men = 7 women (in earning capacity) 1 man = 7/5 women 10 men = 14 women 10 men + 5 women = 14 + 5 = 19 women

7 women earn ₹875 1 woman earns ₹125 19 women earn ₹125 × 19 = ₹2375

Question 17

A and B working together can finish a job in T days. If A alone takes (T+3) days and B alone takes (T+12) days, find T.

Solution: 1/(T+3) + 1/(T+12) = 1/T [(T+12) + (T+3)] / [(T+3)(T+12)] = 1/T (2T + 15)T = (T+3)(T+12) 2T² + 15T = T² + 15T + 36 T² = 36, T = 6

Question 18

Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both pipes are opened together, but pipe A is turned off after some time and the tank is filled in 18 minutes total, after how many minutes was A turned off?

Solution: Let A work for x minutes A's rate = 1/20 per min, B's rate = 1/30 per min

x/20 + 18/30 = 1 x/20 = 1 - 18/30 = 12/30 = 2/5 x = 40/5 = 8 minutes

Question 19

8 men working 9 hours a day complete a piece of work in 20 days. In how many days can 7 men working for 10 hours a day complete the same piece of work?

Solution: Total work = 8 men × 9 hours × 20 days = 1440 man-hours Days needed = 1440 / (7 × 10) = 1440/70 = 144/7 = 20.57 days

Question 20

A can do 1/3 of a work in 5 days and B can do 2/5 of the work in 10 days. In how many days can both A and B together do the work?

Solution: A does 1/3 work in 5 days, so full work in 15 days B does 2/5 work in 10 days, so full work in 25 days

A's 1 day work = 1/15, B's 1 day work = 1/25 Together: 1/15 + 1/25 = (5+3)/75 = 8/75 Time = 75/8 = 9.375 days

Tips & Tricks for Time and Work Problems

1. LCM Method for Quick Calculation

Instead of dealing with fractions, assume total work as LCM of the given days. Example: If A takes 12 days and B takes 18 days, assume work = 36 units. A's efficiency = 3 units/day, B's = 2 units/day. Together = 5 units/day. Time = 36/5 = 7.2 days.

2. Efficiency Ratio Shortcut

If A is twice as efficient as B, and they complete work in x days together:

A takes 3x/2 days alone
B takes 3x days alone

3. Man-Days Concept

Always remember: Men × Days = Constant for the same work If M₁D₁ = M₂D₂, you can find unknown values quickly.

4. Work Left Formula

If A and B work together for n days, work remaining = 1 - n(1/A + 1/B)

5. Three Workers Formula

If (A+B), (B+C), and (A+C) complete work in x, y, z days respectively: A alone takes: 2xyz/(yz + xz - xy) days

6. Leaving/Joining Pattern

When someone leaves or joins mid-way, calculate work done before the change, then solve for remaining work.

7. Wages Distribution

Wages are distributed in the ratio of efficiency (1/time taken). If A takes 10 days and B takes 15 days, wage ratio = 15:10 = 3:2

Common Mistakes to Avoid

❌ Mistake 1: Confusing Efficiency with Time

Many students assume if A is twice as efficient as B, A takes twice the time. Correction: More efficiency means less time. If A is twice as efficient, A takes half the time.

❌ Mistake 2: Adding Times Instead of Rates

Never add time directly: If A takes 10 days and B takes 15 days, together they DON'T take 25 days. Add their work rates (1/10 + 1/15).

❌ Mistake 3: Ignoring Partial Work Days

When calculating "work done in 3 days," ensure you're using the correct fraction of total work.

❌ Mistake 4: Forgetting to Convert Units

If the question mentions hours per day, ensure all calculations use consistent units.

❌ Mistake 5: Misinterpreting "Men and Women" Problems

Always establish the relationship between men's and women's work capacity using the given equations.

❌ Mistake 6: Calculation Errors with Fractions

Use the LCM method to avoid complex fraction calculations and reduce errors.

Conclusion

Time and Work problems are scoring opportunities in placement exams. With the right approach and sufficient practice, you can solve these questions in under a minute. Remember to:

Master the LCM method for faster calculations
Understand the relationship between efficiency and time
Practice different variations of leaving/joining scenarios
Always verify your answer using an alternative method

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Keywords: Time and Work problems, placement aptitude, TCS NQT preparation, campus placement questions, work efficiency problems, aptitude shortcuts, placement exam tips

Frequently Asked Questions

What is the typical salary range for candidates selected through Time & Work problem preparation in placement exams?

Time and Work questions usually don’t directly determine salary; they help you qualify for the aptitude/technical rounds that decide your final offer. In most Indian campus drives (TCS/Infosys/Wipro/Accenture), the salary band is largely based on your overall performance across aptitude, coding/technical, and interviews, not a single topic.

What eligibility is generally required to attempt Time and Work problems for campus placements?

There’s usually no special eligibility for practicing these questions, any student preparing for campus placements can attempt them. However, you should have basic arithmetic and ratio concepts (fractions, LCM/HCF, unitary method) to solve them efficiently.

How difficult are Time and Work problems in placement exams?

They are typically rated medium difficulty, but the difficulty increases when problems include multiple workers, varying efficiencies, or time-lag scenarios. Many placement questions are designed to test speed and accuracy using standard formulas and reasoning rather than advanced math.

What are the best preparation tips to score well in Time and Work sections?

Start by mastering the core idea: work = rate × time, and use “efficiency” as the rate. Practice a mix of basic (single worker) and advanced (two/three workers, alternate work, partial completion) questions, and revise shortcut approaches like using LCM for combined rates.

How many interview rounds typically include aptitude topics like Time and Work?

Most campus processes include an aptitude/quantitative section in either the first online test or an early screening stage. If you clear that, later rounds (technical/coding/interview) may not repeat Time and Work, but your quantitative strength helps you perform better overall.

What common topics are usually paired with Time and Work in placement aptitude tests?

Time and Work is often grouped with Time-Speed-Distance, Pipes & Cisterns, Percentages, Profit & Loss, and Ratio/Proportion. Expect questions that test the same underlying “rate” logic, sometimes disguised with efficiency, combined work, or completion percentages.

How do I apply or register for campus placement tests where Time and Work is relevant?

You typically apply through your college’s placement cell or the company’s campus portal during the recruitment window. After registration, you’ll receive details for the online test/assessment; focus on aptitude preparation (including Time and Work) alongside coding and technical basics.

What is the selection rate, and how much does Time and Work preparation improve my chances?

Selection rate varies widely by company, batch size, and competition, so there’s no single fixed percentage. However, strong aptitude performance can improve your chances of clearing early screening rounds, which is crucial because many candidates get filtered before technical interviews.